Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $p = \dfrac{x + 8}{5x^3 + 35x^2 - 40x} \div \dfrac{x + 8}{-4x^3 + 32x^2 - 28x} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{x + 8}{5x^3 + 35x^2 - 40x} \times \dfrac{-4x^3 + 32x^2 - 28x}{x + 8} $ First factor out any common factors. $p = \dfrac{x + 8}{5x(x^2 + 7x - 8)} \times \dfrac{-4x(x^2 - 8x + 7)}{x + 8} $ Then factor the quadratic expressions. $p = \dfrac {x + 8} {5x(x - 1)(x + 8)} \times \dfrac {-4x(x - 1)(x - 7)} {x + 8} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {(x + 8) \times -4x(x - 1)(x - 7) } { 5x(x - 1)(x + 8) \times (x + 8)} $ $p = \dfrac {-4x(x - 1)(x - 7)(x + 8)} {5x(x - 1)(x + 8)(x + 8)} $ Notice that $(x - 1)$ and $(x + 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-4x\cancel{(x - 1)}(x - 7)(x + 8)} {5x\cancel{(x - 1)}(x + 8)(x + 8)} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $p = \dfrac {-4x\cancel{(x - 1)}(x - 7)\cancel{(x + 8)}} {5x\cancel{(x - 1)}\cancel{(x + 8)}(x + 8)} $ We are dividing by $x + 8$ , so $x + 8 \neq 0$ Therefore, $x \neq -8$ $p = \dfrac {-4x(x - 7)} {5x(x + 8)} $ $ p = \dfrac{-4(x - 7)}{5(x + 8)}; x \neq 1; x \neq -8 $